Transcript Lecture 8
Lecture 8 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Lecture 8 – Tuesday 2/5/2013 Block 1: Block 2: Block 3: Block 4: Mole Balances Rate Laws Stoichiometry Combine Pressure Drop Liquid Phase Reactions Gas Phase Reactions Engineering Analysis of Pressure Drop 2 Pressure Drop in PBRs Concentration Flow System: Gas Phase Flow System: CA 3 FA CA FA 0 1 X FA0 1 X C 1 X T0 P A0 T P0 1 X T P0 0 1 X T0 P b b FA0 B X C A0 B X FB a a T0 P CB P T 1 X T P0 1 X 0 0 T0 P T P0 T0 P Pressure Drop in PBRs Note: Pressure Drop does NOT affect liquid phase reactions Sample Question: Analyze the following second order gas phase reaction that occurs isothermally in a PBR: AB Mole Balances Must use the differential form of the mole balance to separate variables: dX FA0 rA dW Rate Laws Second order in A and irreversible: rA kC A2 4 Pressure Drop in PBRs 1 X P T0 CA CA 0 1 X P0 T FA Stoichiometry Isothermal, T=T0 Combine: 1 X P CA CA 0 1 X P0 dX kC 1 X P 2 dW FA0 1 X P0 2 A0 2 2 Need to find (P/P0) as a function of W (or V if you have a PFR) 5 Pressure Drop in PBRs Ergun Equation: dP G 1 1501 3 1 .75 G dz g c D p Dp TURBULENT LAMINAR Constant mass flow: m m 0 0 0 0 0 6 FT P0 T 0 FT 0 P T0 P0 T 0 (1 X ) P T0 Pressure Drop in PBRs Variable Density dP G dz 0 g c D p Let 7 P T0 FT 0 0 P0 T FT P0 T FT 1 1501 3 1.75G Dp P T0 FT 0 G 0 0 gc Dp 1 1501 3 1.75G Dp Pressure Drop in PBRs Catalyst Weight Where Let 8 W zAc b zAc 1 c b bulk density c solid catalyst density porosity (a.k.a., void fraction ) (1 ) solid fraction 0 P0 T FT dP dW Ac 1 c P T0 FT 0 2 0 1 Ac 1 c P0 Pressure Drop in PBRs dy T FT dW 2 y T0 FT 0 P y P0 We will use this form for single reactions: d P P0 1 T 1 X dW 2 P P0 T0 dy T 1 X dW 2 y T0 9 dy 1 X dW 2y Isothermal case Pressure Drop in PBRs kC 1 X 2 dX y 2 dW FA0 1 X 2 A0 2 dX dP f X , P and f X , P or dW dW dy f y, X dW The two expressions are coupled ordinary differential equations. We can only solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical solution. 10 Polymath will combine the Mole Balances, Rate Laws and Stoichiometry. Packed Bed Reactors For 0 dy (1 X ) dW 2 y When W 0 y 1 dy dW 2 y (1 W ) 2 y (1 W )1/ 2 11 1 Pressure Drop in a PBR P 12 W 2 Concentration Profile in a PBR CA P CA CA 0 1 X P0 No P P W 13 3 Reaction Rate in a PBR -rA P rA kC k (1 X ) P0 2 A 2 No P W 14 P 2 4 Conversion in a PBR X No P P W 15 5 Flow Rate in a PBR P For 0 : f 0 P0 0 P 1 No P W 16 P0 T 0 1 X P T0 T T0 P0 y P 0 1 f (1 X ) y 17 Example 1: Gas Phase Reaction in PBR for =0 Gas Phase reaction in PBR with δ = 0 (Analytical Solution) A + B 2C Repeat the previous one with equimolar feed of A and B and: kA = 1.5dm6/mol/kg/min CA0 CB0 α = 0.0099 kg-1 Find X at 100 kg 18 C A0 CB0 X ? Example 1: Gas Phase Reaction in PBR for =0 1) Mole Balance dX r ' A dW FA0 2) Rate Law r ' A kCACB 3) Stoichiometry CA CA0 1 X y CB CA0 1 X y 19 Example 1: Gas Phase Reaction in PBR for =0 dy dW 2y W 0 2, ydy dW y 2 1 W , y 1 y 1 W 12 4) Combine rA kC 1 X y kC 1 X 1 W 2 A0 2 2 2 A0 dX kCA2 0 1 X 1 W dW FA0 2 20 2 Example 1: Gas Phase Reaction in PBR for =0 kCA2 0 dX 1 W dW 2 1 X FA0 kCA2 0 X W 2 W 1 X FA0 2 W 0, X 0, W W , X X X 0.6 with pressure drop X 0.75 without pressure drop, i.e. 0 21 Example 2: Gas Phase Reaction in PBR for ≠0 The reaction A + 2B C is carried out in a packed bed reactor in which there is pressure drop. The feed is stoichiometric in A and B. Plot the conversion and pressure ratio y = P/P0 as a function of catalyst weight up to 100 kg. Additional Information kA = 6 dm9/mol2/kg/min α = 0.02 kg-1 22 Example 2: Gas Phase Reaction in PBR for ≠0 A + 2B C 1) Mole Balance dX rA dW FA0 2) Rate Law rA kCACB2 3) Stoichiometry: Gas, Isothermal P0 0 1 X P 23 1 X C A C A0 y 1 X Example 2: Gas Phase Reaction in PBR for ≠0 B 2 X 4) CB C A0 1 X y dy 1 X 5) dW 2y 1 X 6) f 0 y 1 2 7) y A0 [1 1 2] [2] 3 3 C A0 2, FA0 2, k 6, 0.02 24 Initial values: W=0, X=0, y=1 Final values: W=100 Combine with Polymath. If δ≠0, polymath must be used to solve. Example 2: Gas Phase Reaction in PBR for ≠0 25 Example 2: Gas Phase Reaction in PBR for ≠0 26 T = T0 27 Robert the Worrier wonders: What if we increase the catalyst size by a factor of 2? Pressure Drop Engineering Analysis 𝜌0 = 𝑀𝑊 ∗ 𝐶𝑇0 = 𝑀𝑊 ∗ 𝑃0 𝑅𝑇0 2𝑅𝑇0 150 1 − 𝜙 𝜇 𝛼= 𝐺 + 1.75𝐺 𝐷𝑃 𝐴𝐶 𝜌𝐶 𝑔𝐶 𝑃02 𝐷𝑃 𝜙 3 𝑀𝑊 1 𝛼≈ 𝑃0 29 2 Pressure Drop Engineering Analysis 30 Pressure Drop Engineering Analysis 31 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 32 End of Lecture 8 33 Pressure Drop - Summary Pressure Drop Liquid Phase Reactions Pressure Drop does not affect concentrations in lquid phase reactions. Gas Phase Reactions Epsilon does not equal to zero d(P)/d(W)=… Polymath will combine with d(X)/d(W) =… for you Epsilon = 0 and isothermal P=f(W) Combine then separate variables (X,W) and integrate Engineering Analysis of Pressure Drop 34 Pressure Change – Molar Flow Rate FT P0 T 0 FT 0 P T0 dP dW A c 1 c FT T 0 FT 0 T0 dy dW yP0 A c 1 c dy FT T dW 2 y FT 0 T0 35 20 P0 A C 1 C Use for heat effects, multiple rxns FT 1 X Isothermal: T = T0 FT 0 dX 1 X dW 2y Example 1: Gas Phase Reaction in PBR for =0 A + B 2C dm6 1 k 1.5 , 0.0099kg m ol kg min Case 1: W 100kg , X ? , P ? Case 2: DP 2 DP1 C A0 CB0 36 , C B 0 C A0 1 , P02 P01 , 2 X ? , P? X ? P? PBR dX FA0 r ' A dW rA kC ACB FA CA y FT C A CB 0 and T T0 y (1 W )1/2 37